Algebra Examples

$- 3 = - 2 y^{3} - 5$

Step 1

Rewrite the equation as $- 2 y^{3} - 5 = - 3$ .

$- 2 y^{3} - 5 = - 3$

Step 2

Move all terms not containing $y$ to the right side of the equation.

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Step 2.1

Add $5$ to both sides of the equation.

$- 2 y^{3} = - 3 + 5$

Step 2.2

Add $- 3$ and $5$ .

$- 2 y^{3} = 2$

Step 3

Subtract $2$ from both sides of the equation.

$- 2 y^{3} - 2 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $- 2$ out of $- 2 y^{3} - 2$ .

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Step 4.1.1

Factor $- 2$ out of $- 2 y^{3}$ .

$- 2 y^{3} - 2 = 0$

Step 4.1.2

Factor $- 2$ out of $- 2$ .

$- 2 y^{3} - 2 \cdot 1 = 0$

Step 4.1.3

Factor $- 2$ out of $- 2 (y^{3}) - 2 (1)$ .

$- 2 (y^{3} + 1) = 0$

Step 4.2

Rewrite $1$ as $1^{3}$ .

$- 2 (y^{3} + 1^{3}) = 0$

Step 4.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = y$ and $b = 1$ .

$- 2 ((y + 1) (y^{2} - y \cdot 1 + 1^{2})) = 0$

Step 4.4

Factor.

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Step 4.4.1

Simplify.

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Step 4.4.1.1

Multiply $- 1$ by $1$ .

$- 2 ((y + 1) (y^{2} - y + 1^{2})) = 0$

Step 4.4.1.2

One to any power is one.

$- 2 ((y + 1) (y^{2} - y + 1)) = 0$

Step 4.4.2

Remove unnecessary parentheses.

$- 2 (y + 1) (y^{2} - y + 1) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y + 1 = 0$

$y^{2} - y + 1 = 0$

Step 6

Set $y + 1$ equal to $0$ and solve for $y$ .

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Step 6.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 7

Set $y^{2} - y + 1$ equal to $0$ and solve for $y$ .

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Step 7.1

Set $y^{2} - y + 1$ equal to $0$ .

$y^{2} - y + 1 = 0$

Step 7.2

Solve $y^{2} - y + 1 = 0$ for $y$ .

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Step 7.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.2.2

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $y$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 7.2.3

Simplify.

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Step 7.2.3.1

Simplify the numerator.

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Step 7.2.3.1.1

Raise $- 1$ to the power of $2$ .

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 7.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 7.2.3.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 7.2.3.1.2.2

Multiply $- 4$ by $1$ .

$y = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 7.2.3.1.3

Subtract $4$ from $1$ .

$y = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 7.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$y = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 7.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$y = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 7.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 7.2.3.2

Multiply $2$ by $1$ .

$y = \frac{1 \pm i \sqrt{3}}{2}$

Step 7.2.4

The final answer is the combination of both solutions.

$y = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 8

The final solution is all the values that make $- 2 (y + 1) (y^{2} - y + 1) = 0$ true.

$y = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$